Question

Solve the initial value problem. Eigenpairs of the coefficient matrices were determined in Exercises 1-10.\(\mathbf{y}^{\prime}=\left[\begin{array}{rr}0 & -9 \\ 1 & 0\end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{l}6 \\ 2\end{array}\right]\)

Short answer

Answer: The particular solution for the given initial value problem is: $$\phi(t)= 2e^{(3it)}\left[\begin{array}{l}3i \\ 1\end{array}\right].$$

Step-by-step solution

Find the Fundamental Matrix Solution

The equation for the fundamental matrix solution, \(\phi(t)\), can be written as: $$\phi(t)=c_1 e^{\lambda_1 t}\mathbf{v}_1 + c_2 e^{\lambda_2 t}\mathbf{v}_2.$$ Substituting the eigenpairs, we have: $$\phi(t)=c_1 e^{(3it)}\left[\begin{array}{r}3i \\ 1\end{array}\right] + c_2 e^{(-3it)}\left[\begin{array}{r}-3i \\ 1\end{array}\right].$$

Evaluate the Initial Condition

To find the constants \(c_1\) and \(c_2\), we need to evaluate the fundamental matrix solution at the given initial condition, \(t=0\). From the initial value problem, we know that: $$\mathbf{y}(0)=\left[\begin{array}{l}6 \\ 2\end{array}\right].$$ Equating \(\phi(0)\) to the initial condition, we have: $$\left[\begin{array}{r}6 \\ 2\end{array}\right] = c_1 \left[\begin{array}{r}3i \\ 1\end{array}\right] + c_2 \left[\begin{array}{r}-3i \\ 1\end{array}\right].$$

Determine the Constants

To find the constants \(c_1\) and \(c_2\), we can solve the system of equations formed from the initial condition: $$6 = 3ic_1 -3ic_2$$$$2 = c_1 + c_2.$$ Solving for \(c_1\) and \(c_2\), we find that \(c_1=2\) and \(c_2=0\).

Find the Particular Solution

Now with the constants determined, we can write the particular solution for the given initial value problem: $$\phi(t)= 2e^{(3it)}\left[\begin{array}{l}3i \\ 1\end{array}\right].$$ In conclusion, the solution to the given initial value problem is: $$\phi(t)= 2e^{(3it)}\left[\begin{array}{l}3i \\ 1\end{array}\right].$$

Key concepts

Eigenpairs

In the study of linear algebra, eigenpairs (which consist of eigenvalues and eigenvectors) are fundamental to understanding the behavior of linear transformations. An eigenvalue, denoted typically by \( \lambda \), is a special scalar that remains constant when multiplied by its corresponding eigenvector, represented as \( \mathbf{v} \).

Mathematically, for a given matrix \( A \), if applying the matrix to the vector \( \mathbf{v} \) scales the vector by the eigenvalue \( \lambda \), then this relationship is written as \( A\mathbf{v} = \lambda\mathbf{v} \). To solve an initial value problem in differential equations, identifying the eigenpairs of the coefficient matrix is crucial. They provide insight into the system's behavior, and allow for the construction of solutions that describe how the system evolves over time.

Fundamental Matrix Solution

When it comes to solving linear systems of differential equations, the fundamental matrix solution is a powerful tool. It is constructed using eigenpairs and represents the general solution for a system of differential equations. In the form \( \phi(t)=c_1 e^{\lambda_1 t}\mathbf{v}_1 + c_2 e^{\lambda_2 t}\mathbf{v}_2 \) and so on for higher dimensions, it incorporates all possible behaviors of the system. Here, \( c_1, c_2 \), etc., are constants determined by initial conditions, \( \lambda_1, \lambda_2 \), etc., are eigenvalues, and \( \mathbf{v}_1, \mathbf{v}_2 \), etc., are the corresponding eigenvectors.

The fundamental matrix solution spans the solution space of the differential equation, effectively capturing the essence of how any initial state will evolve over time according to the dynamics defined by the system.

Particular Solution

The concept of a particular solution to differential equations is related to solving the equation with specific initial conditions. While the general solution incorporates all possible solutions, a particular solution is one that satisfies the initial conditions of the problem. In the context of the fundamental matrix solution, once the eigenvalues and eigenvectors have been used to form the general solution, the constants in that solution are adjusted to match the given initial state of the system; this produces the particular solution for that given problem.

In our exercise's case, this involved matching the coefficients \( c_1 \) and \( c_2 \) to the system's state at \( t=0 \), using the initial condition \( \mathbf{y}(0) \) to find the values of those constants. This process creates a solution tailored to the problem at hand, encapsulating not just any possible state but the actual state given by the initial conditions.

Differential Equations

The realm of differential equations is expansive, encompassing equations that describe how mathematical functions behave when subject to change. These equations can model numerous physical phenomena in engineering, physics, and other fields. A differential equation contains derivatives, which represent rates of change, and solving a differential equation allows one to find a function (or set of functions) that satisfies the relationship outlined in the equation.

An initial value problem is a specific type of differential equation in which the solution must fulfill conditions at the outset—hence the term 'initial.' By employing methods such as the one demonstrated in the fundamental matrix solution, we can solve these equations to predict the future behavior of complex systems. Understanding how to manipulate and solve differential equations is a cornerstone of applied mathematics and engineering.